Question

Use the appropriate tables to calculate \(\Delta H^{\circ}\) for (a) the reaction between copper(II) oxide and carbon monoxide to give copper metal and carbon dioxide. (b) the decomposition of one mole of methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) to methane and oxygen gases.

Short answer

Question: Calculate the standard enthalpy change for the following reactions: (a) CuO(s) + CO(g) -> Cu(s) + CO2(g) (b) CH3OH(g) -> CH4(g) + 1/2 O2(g) Answer: (a) The standard enthalpy change for the reaction CuO(s) + CO(g) -> Cu(s) + CO2(g) is -127.0 kJ. (b) The standard enthalpy change for the decomposition of one mole of methyl alcohol (CH3OH) to methane and oxygen gases is 126.2 kJ.

Step-by-step solution

Look up the standard enthalpies of formation for the species involved.

Using a table of standard enthalpies of formation, obtain the following values: \(\Delta H^{\circ}_{f}(\mathrm{CuO}) = -156.0\,\mathrm{kJ \cdot mol^{-1}}\) \(\Delta H^{\circ}_{f}(\mathrm{CO}) = -110.5\,\mathrm{kJ \cdot mol^{-1}}\) \(\Delta H^{\circ}_{f}(\mathrm{Cu}) = 0\,\mathrm{kJ \cdot mol^{-1}}\) \(\Delta H^{\circ}_{f}(\mathrm{CO}_{2}) = -393.5\,\mathrm{kJ \cdot mol^{-1}}\)

Calculate the standard enthalpy change for the reaction.

Apply the equation mentioned above: \(\Delta H^{\circ}_{rxn} = [1\times\Delta H^{\circ}_{f}(\mathrm{CO}_{2})+ 1\times\Delta H^{\circ}_{f}(\mathrm{Cu})] - [1\times\Delta H^{\circ}_{f}(\mathrm{CuO})+ 1\times\Delta H^{\circ}_{f}(\mathrm{CO})]\) \(\Delta H^{\circ}_{rxn} = [(-393.5) + (0)] - [(-156.0) + (-110.5)]\) \(\Delta H^{\circ}_{rxn} = -393.5 + 266.5\) \(\Delta H^{\circ}_{rxn} = -127.0\,\mathrm{kJ}\) (b) Calculate the standard enthalpy change for the decomposition of one mole of methyl alcohol to methane and oxygen gases. The balanced chemical equation for this reaction is: CH3OH(g) -> CH4(g) + \(\frac{1}{2}\)O2(g)

Look up the standard enthalpies of formation for the species involved.

Using a table of standard enthalpies of formation, obtain the following values: \(\Delta H^{\circ}_{f}(\mathrm{CH}_{3}\mathrm{OH}) = -201.0\,\mathrm{kJ \cdot mol^{-1}}\) \(\Delta H^{\circ}_{f}(\mathrm{CH}_{4}) = -74.8\,\mathrm{kJ \cdot mol^{-1}}\) \(\Delta H^{\circ}_{f}(\mathrm{O}_{2}) = 0\,\mathrm{kJ \cdot mol^{-1}}\)

Calculate the standard enthalpy change for the reaction.

Apply the equation mentioned above: \(\Delta H^{\circ}_{rxn} = [1\times\Delta H^{\circ}_{f}(\mathrm{CH}_{4})+ 0.5\times\Delta H^{\circ}_{f}(\mathrm{O}_{2})] - [1\times\Delta H^{\circ}_{f}(\mathrm{CH}_{3}\mathrm{OH})]\) \(\Delta H^{\circ}_{rxn} = [(-74.8) + 0.5\times(0)] - [(-201.0)]\) \(\Delta H^{\circ}_{rxn} = -74.8 + 201.0\) \(\Delta H^{\circ}_{rxn} = 126.2\,\mathrm{kJ}\)

Key concepts

Enthalpy Change Calculation

Understanding enthalpy change calculation is fundamental in grasping the heat exchanged during a chemical reaction under constant pressure. Enthalpy, symbolized by the letter H, is a measure of the total energy in a thermodynamic system. It's the sum of the system's internal energy plus the product of its pressure and volume.

To calculate the change in enthalpy for a reaction, we use the standard enthalpies of formation of the reactants and the products. The standard enthalpy of formation, denoted as \( \Delta H^{\circ}_{f} \), is the heat change that results when one mole of a compound is formed from its elements in their standard states at 1 atmosphere of pressure and 25°C (298 K).

The enthalpy change of a reaction, \( \Delta H^{\circ}_{rxn} \) for the overall process is determined by the expression:
\[ \Delta H^{\circ}_{rxn} = \sum (\Delta H^{\circ}_{f} \text{ of products}) - \sum (\Delta H^{\circ}_{f} \text{ of reactants}) \]
This equation suggests that we subtract the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. It's essential to consider the stoichiometry of the reaction when calculating these sums; one must account for the moles of each substance as indicated in the balanced chemical equation.

For instance, in a simple reaction where one mole of copper(II) oxide reacts with carbon monoxide to form copper metal and carbon dioxide, the calculation takes into account the formation enthalpies of each substance and the reaction's stoichiometry.

Thermochemical Reactions

Thermochemical reactions involve changes in the heat content of reactants and products. They are often accompanied by the release or absorption of heat. When a reaction releases heat, it is exothermic, while a reaction that absorbs heat from its surroundings is endothermic.

Understanding thermochemical reactions helps predict the temperature changes that occur during various chemical processes. This insight is crucial in fields such as energy production, environmental science, and materials engineering. Furthermore, it's helpful when considering safety in chemical reactions, as exothermic reactions can release a significant amount of heat that needs to be controlled.

For instance, the reaction between copper(II) oxide and carbon monoxide is exothermic, which we can tell from the negative sign of the \( \Delta H^{\circ}_{rxn} \). This denotes heat is being released to the surroundings. On the other hand, the decomposition of methyl alcohol into methane and oxygen is endothermic, as the positive \( \Delta H^{\circ}_{rxn} \) indicates the system absorbs heat.

Chemical Thermodynamics

Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It involves not just the study of energy changes, but also the spontaneity of processes and the equilibrium of chemical and physical changes.

One key concept in chemical thermodynamics is Gibbs free energy, which combines the concepts of enthalpy, entropy (disorder), and temperature to predict the spontaneity of a process. However, for many practical purposes such as calculating energy changes in reactions, we focus on enthalpy.

Enthalpy changes tell us a lot about a reaction, including how much energy is available to do work and whether a reaction will be product or reactant favored under standard conditions. If a reaction releases energy (exothermic), it may be more likely to occur spontaneously. In contrast, if it absorbs energy (endothermic), other factors, such as entropy, might play a significant role in determining its spontaneity.

By applying the principles of chemical thermodynamics, we can tailor reactions to be more efficient, design better energy storage systems, manufacture products safely, and understand the environmental impact of chemical processes.